3.12 \(\int \cot ^6(c+d x) (a+i a \tan (c+d x)) \, dx\)

Optimal. Leaf size=100 \[ -\frac {a \cot ^5(c+d x)}{5 d}-\frac {i a \cot ^4(c+d x)}{4 d}+\frac {a \cot ^3(c+d x)}{3 d}+\frac {i a \cot ^2(c+d x)}{2 d}-\frac {a \cot (c+d x)}{d}+\frac {i a \log (\sin (c+d x))}{d}-a x \]

[Out]

-a*x-a*cot(d*x+c)/d+1/2*I*a*cot(d*x+c)^2/d+1/3*a*cot(d*x+c)^3/d-1/4*I*a*cot(d*x+c)^4/d-1/5*a*cot(d*x+c)^5/d+I*
a*ln(sin(d*x+c))/d

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Rubi [A]  time = 0.13, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3529, 3531, 3475} \[ -\frac {a \cot ^5(c+d x)}{5 d}-\frac {i a \cot ^4(c+d x)}{4 d}+\frac {a \cot ^3(c+d x)}{3 d}+\frac {i a \cot ^2(c+d x)}{2 d}-\frac {a \cot (c+d x)}{d}+\frac {i a \log (\sin (c+d x))}{d}-a x \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^6*(a + I*a*Tan[c + d*x]),x]

[Out]

-(a*x) - (a*Cot[c + d*x])/d + ((I/2)*a*Cot[c + d*x]^2)/d + (a*Cot[c + d*x]^3)/(3*d) - ((I/4)*a*Cot[c + d*x]^4)
/d - (a*Cot[c + d*x]^5)/(5*d) + (I*a*Log[Sin[c + d*x]])/d

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 0]

Rubi steps

\begin {align*} \int \cot ^6(c+d x) (a+i a \tan (c+d x)) \, dx &=-\frac {a \cot ^5(c+d x)}{5 d}+\int \cot ^5(c+d x) (i a-a \tan (c+d x)) \, dx\\ &=-\frac {i a \cot ^4(c+d x)}{4 d}-\frac {a \cot ^5(c+d x)}{5 d}+\int \cot ^4(c+d x) (-a-i a \tan (c+d x)) \, dx\\ &=\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^4(c+d x)}{4 d}-\frac {a \cot ^5(c+d x)}{5 d}+\int \cot ^3(c+d x) (-i a+a \tan (c+d x)) \, dx\\ &=\frac {i a \cot ^2(c+d x)}{2 d}+\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^4(c+d x)}{4 d}-\frac {a \cot ^5(c+d x)}{5 d}+\int \cot ^2(c+d x) (a+i a \tan (c+d x)) \, dx\\ &=-\frac {a \cot (c+d x)}{d}+\frac {i a \cot ^2(c+d x)}{2 d}+\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^4(c+d x)}{4 d}-\frac {a \cot ^5(c+d x)}{5 d}+\int \cot (c+d x) (i a-a \tan (c+d x)) \, dx\\ &=-a x-\frac {a \cot (c+d x)}{d}+\frac {i a \cot ^2(c+d x)}{2 d}+\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^4(c+d x)}{4 d}-\frac {a \cot ^5(c+d x)}{5 d}+(i a) \int \cot (c+d x) \, dx\\ &=-a x-\frac {a \cot (c+d x)}{d}+\frac {i a \cot ^2(c+d x)}{2 d}+\frac {a \cot ^3(c+d x)}{3 d}-\frac {i a \cot ^4(c+d x)}{4 d}-\frac {a \cot ^5(c+d x)}{5 d}+\frac {i a \log (\sin (c+d x))}{d}\\ \end {align*}

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Mathematica [C]  time = 0.35, size = 84, normalized size = 0.84 \[ -\frac {a \cot ^5(c+d x) \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-\tan ^2(c+d x)\right )}{5 d}+\frac {i a \left (-\cot ^4(c+d x)+2 \cot ^2(c+d x)+4 \log (\tan (c+d x))+4 \log (\cos (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^6*(a + I*a*Tan[c + d*x]),x]

[Out]

-1/5*(a*Cot[c + d*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, -Tan[c + d*x]^2])/d + ((I/4)*a*(2*Cot[c + d*x]^2 - Cot
[c + d*x]^4 + 4*Log[Cos[c + d*x]] + 4*Log[Tan[c + d*x]]))/d

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fricas [B]  time = 0.42, size = 196, normalized size = 1.96 \[ \frac {-150 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 300 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 400 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 200 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (15 i \, a e^{\left (10 i \, d x + 10 i \, c\right )} - 75 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} + 150 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 150 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 75 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 15 i \, a\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right ) - 46 i \, a}{15 \, {\left (d e^{\left (10 i \, d x + 10 i \, c\right )} - 5 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, d e^{\left (6 i \, d x + 6 i \, c\right )} - 10 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/15*(-150*I*a*e^(8*I*d*x + 8*I*c) + 300*I*a*e^(6*I*d*x + 6*I*c) - 400*I*a*e^(4*I*d*x + 4*I*c) + 200*I*a*e^(2*
I*d*x + 2*I*c) + (15*I*a*e^(10*I*d*x + 10*I*c) - 75*I*a*e^(8*I*d*x + 8*I*c) + 150*I*a*e^(6*I*d*x + 6*I*c) - 15
0*I*a*e^(4*I*d*x + 4*I*c) + 75*I*a*e^(2*I*d*x + 2*I*c) - 15*I*a)*log(e^(2*I*d*x + 2*I*c) - 1) - 46*I*a)/(d*e^(
10*I*d*x + 10*I*c) - 5*d*e^(8*I*d*x + 8*I*c) + 10*d*e^(6*I*d*x + 6*I*c) - 10*d*e^(4*I*d*x + 4*I*c) + 5*d*e^(2*
I*d*x + 2*I*c) - d)

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giac [B]  time = 1.89, size = 186, normalized size = 1.86 \[ \frac {6 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 70 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 180 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1920 i \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right ) + 960 i \, a \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + 660 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {-2192 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 660 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 180 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 70 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 15 i \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c)),x, algorithm="giac")

[Out]

1/960*(6*a*tan(1/2*d*x + 1/2*c)^5 - 15*I*a*tan(1/2*d*x + 1/2*c)^4 - 70*a*tan(1/2*d*x + 1/2*c)^3 + 180*I*a*tan(
1/2*d*x + 1/2*c)^2 - 1920*I*a*log(tan(1/2*d*x + 1/2*c) + I) + 960*I*a*log(tan(1/2*d*x + 1/2*c)) + 660*a*tan(1/
2*d*x + 1/2*c) + (-2192*I*a*tan(1/2*d*x + 1/2*c)^5 - 660*a*tan(1/2*d*x + 1/2*c)^4 + 180*I*a*tan(1/2*d*x + 1/2*
c)^3 + 70*a*tan(1/2*d*x + 1/2*c)^2 - 15*I*a*tan(1/2*d*x + 1/2*c) - 6*a)/tan(1/2*d*x + 1/2*c)^5)/d

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maple [A]  time = 0.34, size = 97, normalized size = 0.97 \[ -\frac {i a \left (\cot ^{4}\left (d x +c \right )\right )}{4 d}+\frac {i a \left (\cot ^{2}\left (d x +c \right )\right )}{2 d}+\frac {i a \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {a \left (\cot ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a \left (\cot ^{3}\left (d x +c \right )\right )}{3 d}-\frac {a \cot \left (d x +c \right )}{d}-a x -\frac {c a}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^6*(a+I*a*tan(d*x+c)),x)

[Out]

-1/4*I*a*cot(d*x+c)^4/d+1/2*I*a*cot(d*x+c)^2/d+I*a*ln(sin(d*x+c))/d-1/5*a*cot(d*x+c)^5/d+1/3*a*cot(d*x+c)^3/d-
a*cot(d*x+c)/d-a*x-1/d*c*a

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maxima [A]  time = 0.78, size = 93, normalized size = 0.93 \[ -\frac {60 \, {\left (d x + c\right )} a + 30 i \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 60 i \, a \log \left (\tan \left (d x + c\right )\right ) + \frac {60 \, a \tan \left (d x + c\right )^{4} - 30 i \, a \tan \left (d x + c\right )^{3} - 20 \, a \tan \left (d x + c\right )^{2} + 15 i \, a \tan \left (d x + c\right ) + 12 \, a}{\tan \left (d x + c\right )^{5}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^6*(a+I*a*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/60*(60*(d*x + c)*a + 30*I*a*log(tan(d*x + c)^2 + 1) - 60*I*a*log(tan(d*x + c)) + (60*a*tan(d*x + c)^4 - 30*
I*a*tan(d*x + c)^3 - 20*a*tan(d*x + c)^2 + 15*I*a*tan(d*x + c) + 12*a)/tan(d*x + c)^5)/d

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mupad [B]  time = 4.07, size = 79, normalized size = 0.79 \[ -\frac {2\,a\,\mathrm {atan}\left (2\,\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{d}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^4-\frac {1{}\mathrm {i}\,a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{2}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{3}+\frac {1{}\mathrm {i}\,a\,\mathrm {tan}\left (c+d\,x\right )}{4}+\frac {a}{5}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^6*(a + a*tan(c + d*x)*1i),x)

[Out]

- (2*a*atan(2*tan(c + d*x) + 1i))/d - (a/5 + (a*tan(c + d*x)*1i)/4 - (a*tan(c + d*x)^2)/3 - (a*tan(c + d*x)^3*
1i)/2 + a*tan(c + d*x)^4)/(d*tan(c + d*x)^5)

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sympy [B]  time = 0.58, size = 206, normalized size = 2.06 \[ \frac {i a \log {\left (e^{2 i d x} - e^{- 2 i c} \right )}}{d} + \frac {- 150 i a e^{8 i c} e^{8 i d x} + 300 i a e^{6 i c} e^{6 i d x} - 400 i a e^{4 i c} e^{4 i d x} + 200 i a e^{2 i c} e^{2 i d x} - 46 i a}{15 d e^{10 i c} e^{10 i d x} - 75 d e^{8 i c} e^{8 i d x} + 150 d e^{6 i c} e^{6 i d x} - 150 d e^{4 i c} e^{4 i d x} + 75 d e^{2 i c} e^{2 i d x} - 15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**6*(a+I*a*tan(d*x+c)),x)

[Out]

I*a*log(exp(2*I*d*x) - exp(-2*I*c))/d + (-150*I*a*exp(8*I*c)*exp(8*I*d*x) + 300*I*a*exp(6*I*c)*exp(6*I*d*x) -
400*I*a*exp(4*I*c)*exp(4*I*d*x) + 200*I*a*exp(2*I*c)*exp(2*I*d*x) - 46*I*a)/(15*d*exp(10*I*c)*exp(10*I*d*x) -
75*d*exp(8*I*c)*exp(8*I*d*x) + 150*d*exp(6*I*c)*exp(6*I*d*x) - 150*d*exp(4*I*c)*exp(4*I*d*x) + 75*d*exp(2*I*c)
*exp(2*I*d*x) - 15*d)

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